PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 98(112) (2015), 199–210 DOI: 10.2298/PIM141009007A ON THE GROWTH AND THE ZEROS OF SOLUTIONS OF HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH MEROMORPHIC COEFFICIENTS Maamar Andasmas and Benharrat Belaïdi Abstract. We investigate the growth of meromorphic solutions of homogeneous and nonhomogeneous higher order linear differential equations f (k) + k−1 X Aj f (j) + A0 f = 0 (k > 2), j=1 k−1 f (k) + X Aj f (j) + A0 f = Ak (k > 2), j=1 where Aj (z) (j = 0, 1, . . . , k) are meromorphic functions with finite order. Under some conditions on the coefficients, we show that all meromorphic solutions f 6≡ 0 of the above equations have an infinite order and infinite lower order. Furthermore, we give some estimates of their hyper-order, exponent and hyper-exponent of convergence of distinct zeros. We improve the results due to Kwon; Chen and Yang; Belaïdi; Chen; Shen and Xu. 1. Introduction and statement of results In this paper, we use the standard notations of Nevanlinna’s value distribution theory [10, 12, 15]. In addition, we use the notations λ(f ) and λ(1/f ) to denote respectively the exponents of convergence of the zeros and the poles of a meromorphic function f , ρ(f ) and µ(f ) to denote respectively the order and the lower order of f . The hyper-order ρ2 (f ), the hyper-exponent λ2 (f ) of convergence of zeros and the hyper-exponent λ2 (f ) of convergence of distinct zeros of f are defined respectively by log log T (r, f ) log log N (r, 1/f ) ρ2 (f ) = lim sup , λ2 (f ) = lim sup , log r log r r→+∞ r→+∞ 2010 Mathematics Subject Classification: Primary 34M10; Secondary 30D35. Key words and phrases: linear differential equations; meromorphic functions; order of growth; hyper-order; exponent of convergence of zeros; hyper-exponent of convergence of zeros. Partially supported by University of Mostaganem (UMAB) (CNEPRU Project Code B02220120024). Communicated by Gradimir Milovanović. 199 200 ANDASMAS AND BELAÏDI log log N (r, 1/f ) . log r r→+∞ First, we recall the following definitions. The linear measure of a set E ⊂ (0, +∞) is R +∞ defined as m(E) = 0 χE (t) dt and the logarithmic measure of a set F ⊂ (1, +∞) R +∞ 1 is defined by lm(F ) = 1 t χF (t) dt, where χH (t) is the characteristic function of a set H. The upper density of a set E ⊂ (0, +∞) is defined by λ2 (f ) = lim sup dens E = lim sup r→+∞ m(E ∩ [0, r]) . r The upper logarithmic density of a set F ⊂ (1, +∞) is defined by log dens(F ) = lim sup r→+∞ lm(F ∩ [1, r]) . log r Proposition 1.1. For all H ⊂ [1, +∞) the following statements hold: i) If lm(H) = ∞, then m(H) = ∞; ii) If dens H > 0, then m(H) = ∞; iii) If log dens H > 0, then lm(H) = ∞. Proof. i) Since we have χH (t)/t 6 χH (t) for all t ∈ H ⊂ [1, +∞), then m(H) > lm(H). We can easily prove the results ii) and iii) by applying the definition of the limit and the properties m(H ∩ [0, r]) 6 m(H) and lm(H ∩ [1, r]) 6 lm(H). In [11], Kwon investigated the growth of second order equations and obtained the following result. Theorem 1.1. [11] Let H be a set of complex numbers satisfying dens{|z| : z ∈ H} > 0, and let A(z) and B(z) be entire functions such that for real constants α (> 0), β (> 0), |A(z)| 6 exp{o(1)|z|β } and |B(z)| > exp{(1 + o(1))α|z|β } as z → ∞ for z ∈ H. Then every solution f 6≡ 0 of the equation (1.1) f ′′ + A(z)f ′ + B(z)f = 0 has infinite order and ρ2 (f ) > β. In [6], Chen and Yang have studied the growth of solutions of (1.1) and obtained the following result. Theorem 1.2. [6] Let H be a set of complex numbers satisfying dens{|z| : z ∈ H} > 0, and let A(z) and B(z) be entire functions with ρ(A) 6 ρ(B) = ρ < +∞ such that for real constant C (> 0) and for any given ε > 0, |A(z)| 6 exp{o(1)|z|ρ−ε } and |B(z)| > exp{(1 + o(1))C|z|ρ−ε } as z → ∞ for z ∈ H. Then every solution f 6≡ 0 of equation (1.1) has infinite order and ρ2 (f ) = ρ(B). ON THE GROWTH AND THE ZEROS OF SOLUTIONS 201 These results were improved by Belaïdi in [1, 2] by considering more general conditions to higher order linear differential equations with entire coefficients. Recently in [8], Chen extended the previous results by studying the zeros and the growth of meromorphic solutions of equation (1.1) and the non-homogeneous equation f ′′ + A(z)f ′ + B(z)f = F when A(z), B(z), F (z) are meromorphic functions. Here we consider for k > 2 the homogeneous and the non-homogeneous linear differential equations (1.2) f (k) + k−1 X Aj f (j) + A0 f = 0, j=1 (1.3) f (k) + k−1 X Aj f (j) + A0 f = Ak , j=1 where Aj (z) (j = 0, 1, . . . , k) (A0 6≡ 0 and Ak 6≡ 0) are meromorphic functions with finite order. We investigate the zeros and growth of meromorphic solutions of equations (1.2) and (1.3). The present article may be understood as an extension and improvement of Theorems 2.3 and 2.4 in the paper of Shen and Xu [14]. We improve the results due to Chen; Shen and Xu greatly and we give two corollaries in the case when ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} < ρ(A0 ) = σ < 1/2. Theorem 1.3. Let H ⊂ [0, +∞) be a set with infinite linear measure, and let Aj (z) (j = 0, 1, . . . , k − 1) be meromorphic functions with finite order. If there exist positive constants σ > 0, α > 0 such that ρ = max{ρ(Aj ) : j = 1, 2, . . . , k − 1} < σ σ and |A0 (z)| > eαr as |z| = r ∈ H, r → +∞, then every meromorphic solution f 6≡ 0 of equation (1.2) satisfies µ(f ) = ρ(f ) = ∞ and ρ2 (f ) > σ. Furthermore, if λ(1/f ) < ∞, then σ 6 ρ2 (f ) 6 ρ(A0 ). Remark 1.1. By using Proposition 1.1, we can obtain the same results in Theorem 1.3, while putting H ⊂ [0, +∞) to be a set with positive upper density (or while putting H ⊂ [1, +∞) to be a set with positive upper logarithmic density) (or while putting H ⊂ [1, +∞) to be a set with infinite logarithmic measure) instead of putting H to be a set with infinite linear measure. Theorem 1.4. Let H ⊂ [0, +∞) be a set with a positive upper density, and let Aj (z) (j = 0, 1, . . . , k) (Ak 6≡ 0) be meromorphic functions with finite order. If there exist positive constants σ > 0, α > 0 such that ρ = max{ρ(Aj ) : σ j = 1, 2, . . . , k} < σ and |A0 (z)| > eαr as |z| = r ∈ H, r → +∞, then every meromorphic solution f with λ(1/f ) < σ of equation (1.3) is of infinite order and λ(f ) = λ(f ) = ρ(f ) = ∞, λ2 (f ) = λ2 (f ) = ρ2 (f ). Furthermore, if λ(1/f ) < min{µ(f ), σ}, then ρ2 (f ) 6 ρ(A0 ). Remark 1.2. It is clear that ρ(A0 ) = β > σ in Theorems 1.3 and 1.4. Indeed, suppose that ρ(A0 ) = β < σ. Then, by using Lemma 2.2 of this paper, there exists a set E1 ⊂ (1, +∞) that has a finite linear measure such that when |z| = r ∈ / 202 ANDASMAS AND BELAÏDI [0, 1] ∪ E1 , r → +∞, we have for any given ε (0 < ε < σ − β) (1.4) |A0 (z)| 6 er β+ε . On the other hand, by the hypotheses of Theorems 1.3 and 1.4, there exist positive constants σ > 0, α > 0 such that (1.5) |A0 (z)| > eαr σ as |z| = r ∈ H, r → +∞, where H is a set with m(H) = ∞. From (1.4) and (1.5), we obtain for |z| = r ∈ H r [0, 1] ∪ E1 , r → +∞ σ eαr 6 |A0 (z)| 6 er β+ε and by ε (0 < ε < σ − β) this is a contradiction as r → +∞. Hence ρ(A0 ) = β > σ. Corollary 1.1. Let Aj (z) (j = 0, 1, . . . , k) (Ak 6≡ 0) be meromorphic functions having only finitely many poles such that ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} < ρ(A0 ) = σ < 1/2. Then every meromorphic solution f 6≡ 0 of equation (1.2) satisfies µ(f ) = ρ(f ) = ∞ and ρ2 (f ) = σ. Furthermore, every meromorphic solution f of equation (1.3) satisfies λ(f ) = λ(f ) = ρ(f ) = ∞, λ2 (f ) = λ2 (f ) = ρ2 (f ) and if λ(1/f ) < µ(f ), then ρ2 (f ) 6 ρ(A0 ). Corollary 1.2. Let Aj (z) (j = 0, 1, . . . , k) (Ak 6≡ 0) be meromorphic functions such that λ(1/A0 ) < µ(A0 ) 6 ρ(A0 ) = σ < 1/2 and ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} < σ. Then every meromorphic solution f 6≡ 0 of equation (1.2) whose all poles are of uniformly bounded multiplicity satisfies µ(f ) = ρ(f ) = ∞ and ρ2 (f ) = σ. Furthermore, every meromorphic solution f of equation (1.3) whose all poles are of uniformly bounded multiplicity satisfies λ(f ) = λ(f ) = ρ(f ) = ∞, λ2 (f ) = λ2 (f ) = ρ2 (f ) and if λ(1/f ) < µ(f ), then ρ2 (f ) 6 ρ(A0 ). 2. Auxiliary lemmas Lemma 2.1. [9] Let f (z) be a transcendental meromorphic function, and let α > 1, ε > 0 be given constants. Then there exists a set E0 ⊂ [0, ∞) that has finite linear measure and there exists a constant c > 0, such that for all z satisfying |z| = r ∈ / E0 , we have f (j) (z) 6 c[T (αr, f )rε log T (αr, f )]j (j ∈ N). f (z) Lemma 2.2. [5] Let f (z) be a meromorphic function of order ρ(f ) = ρ < +∞. Then for any given ε > 0, there exists a set E1 ⊂ (1, +∞) that has finite linear measure and finite logarithmic measure such that when |z| = r ∈ / [0, 1] ∪ E1 , r → +∞, we have |f (z)| 6 exp{rρ+ε }. P∞ Let g(z) = n=0 an z n be an entire function. We define by µ(r) = max{|an |rn ; n = 0, 1, · · · } the maximum term of g, and define by νg (r) = max{m; µ(r) = |am |rm } the central index of g. ON THE GROWTH AND THE ZEROS OF SOLUTIONS 203 Lemma 2.3. [7] Let f (z) = g(z)/d(z) be a meromorphic function, where g(z) and d(z) are entire functions such that µ(g) = µ(f ) = µ 6 ρ(g) = ρ(f ) 6 +∞, λ(d) = ρ(d) = λ(1/f ) = β < µ. Let z be a point with |z| = r at which |g(z)| = M (r, g) and νg (r) denote the central index of g(z). Then there exists a set E2 ⊂ (1, +∞) with finite logarithmic measure lm(E2 ) < ∞, such that for all z satisfying |z| = r ∈ / [0, 1] ∪ E2 , we have f (n) (z) νg (r) n = (1 + o(1)) (n > 1 is an integer). f (z) z Lemma 2.4. [6] Let g(z) be an entire function of infinite order, with the hyperorder ρ2 (g) = σ. Then log log νg (r) lim sup = σ. log r r→+∞ Lemma 2.5. [13] Let g(z) be an entire function of infinite order. Denote M (r, g) = max{|g(z)| : |z| = r}, then for any sufficiently large number λ > 0, and any r ∈ H0 ⊂ (1, +∞) M (r, g) > c1 exp{c2 rλ }, where lm(H0 ) = ∞ and c1 , c2 are positive constants. Lemma 2.6. Suppose that k > 2 and h0 , h1 , . . . , hk (h0 6≡ 0) are meromorphic functions. Let ρ = max{ρ(hj ) : j = 0, 1, . . . , k} < ∞ and let f be a meromorphic solution of infinite order of the equation (2.1) f (k) + hk−1 f (k−1) + · · · + h0 f = hk with λ(1/f ) = µ < µ(f ). Then ρ2 (f ) 6 ρ. Proof. We assume that f is a meromorphic solution of infinite order ρ(f ) = ∞ of equation (2.1). We can rewrite (2.1) as (s) (k) k−1 X f hk f + . (2.2) |hs | f 6 |h0 | + f f s=1 By Hadamard factorization theorem, we can write f as f (z) = and d(z) are entire functions such that g(z) d(z) , where g(z) µ(f ) = µ(g) 6 ρ(f ) = ρ(g) = ∞, λ(d) = ρ(d) = λ(1/f ) = µ < µ(f ) ρ2 (f ) = ρ2 (g). By Lemma 2.3, there exists a set E2 ⊂ (1, +∞) with finite logarithmic measure, such that for all z satisfying |z| = r ∈ / [0, 1] ∪ E2 at which |g(z)| = M (r, g), we have (n) f (z) νg (r) n (2.3) = (1 + o(1)) (n > 1). f (z) z 204 ANDASMAS AND BELAÏDI Since ρ(hk (z)d(z)) 6 ρ1 = max{µ, ρ}, then by Lemma 2.2, for any ε > 0 there exists a set E1 ⊂ (1, +∞) with a finite logarithmic measure such that (2.4) |hk (z)d(z)| 6 er ρ1 +ε holds for all z satisfying |z| = r ∈ / [0, 1] ∪ E1 , r → +∞. On the other hand, by Lemma 2.5 for a sufficiently large number λ > ρ1 , there exist a set H0 ⊂ (1, +∞) with lm(H0 ) = ∞ and positive constants c1 , c2 such that for any r ∈ H0 (2.5) M (r, g) > c1 exp{c2 rλ }. By (2.4) and (2.5), for any given ε with 0 < ε < λ − ρ1 and for all z satisfying |z| = r ∈ H0 r ([0, 1] ∪ E1 ), r → +∞ at which |g(z)| = M (r, g), we have ρ1 +ε h (z) d(z)h (z) er k k → 0, r → +∞. (2.6) = 6 f (z) g(z) c1 e c 2 r λ Since ρ = max{ρ (hj ) : j = 0, 1, . . . , k} < ∞, then by Lemma 2.2, we have (2.7) |hj (z)| 6 er ρ+ε (j = 0, 1, . . . , k) holds for all z satisfying |z| = r ∈ / [0, 1] ∪ E1 , r → +∞. Substituting (2.3), (2.6) and (2.7) into (2.2), we obtain for all z satisfying |z| = r ∈ H0 r ([0, 1] ∪ E1 ∪ E2 ), r → +∞ at which |g(z)| = M (r, g), k−1 ν (r) k X ρ+ε νg (r) s g r ρ+ε + er |1 + o(1)| 6 e |1 + o(1)| + o(1). z z s=1 So, we get (2.8) |νg (r)|k |1 + o(1)| 6 (k + 1)rk er ρ+ε |νg (r)|k−1 |1 + o(1)|. Then, by Lemma 2.4, we obtain from (2.8) that ρ2 (g) = ρ2 (f ) 6 ρ + ε. Since ε (0 < ε < λ − ρ1 ) being arbitrary, then we get ρ2 (f ) 6 ρ. Lemma 2.7. [3] Let f (z) be an entire function of order ρ, where 0 < ρ(f ) = ρ < 1/2, and let ε > 0 be a given constant. Then there exists a set H1 ⊂ [0, +∞) with dens H1 > 1−2ρ such that |f (z)| > exp{rρ−ε } for all z satisfying |z| = r ∈ H1 . Lemma 2.8. [7] Suppose that h(z) is a meromorphic function with λ(1/h) < µ(h) 6 ρ(h) = σ < 1/2. Then for any ε > 0, there exists a set H2 ⊂ (1, +∞) that has a positive upper logarithmic density such that for all z satisfying |z| = r ∈ H2 , we have |h(z)| > exp{(1 + o(1))rσ−ε }. Lemma 2.9. [11] Let g(z) be a nonconstant entire function of finite order. Then for any given ε > 0, there exists a set H3 ⊂ [0, +∞) with dens H3 = 1 such that M (r, g) > exp{rρ(g)−ε } for all z satisfying |z| = r ∈ H3 . Lemma 2.10. [4] Let Aj (j = 0, 1, . . . , k−1), F 6≡ 0 be finite order meromorphic functions. If f (z) is an infinite order meromorphic solution of the equation f (k) + Ak−1 f (k−1) + · · · + A1 f ′ + A0 f = F, then f satisfies λ(f ) = λ(f ) = ρ(f ) = ∞. ON THE GROWTH AND THE ZEROS OF SOLUTIONS 205 Lemma 2.11. Let H ⊂ [0, +∞) be a set with a positive upper density (or of infinite linear measure), and let hj (z) (j = 0, 1, . . . , k) (h0 6≡ 0) be meromorphic functions with finite order. If there exist positive constants σ > 0, α > 0 such that σ |h0 (z)| > eαr as |z| = r ∈ H, r → +∞, and ρ = max{ρ(hj ) : j = 1, 2, . . . , k} < σ, then every meromorphic solution f 6≡ 0 of equation (2.9) f (k) + k−1 X hj f (j) + h0 f = hk (k > 2), j=1 is transcendental and satisfies ρ(f ) > σ. Proof. Assume that f 6≡ 0 is a meromorphic solution of (2.9) with ρ(f ) < σ. It follows from (2.9) that k−1 (2.10) f (k) X f (j) hk − − hj = h0 . f f f j=1 Since ρ(hj ) < σ (j = 1, 2, . . . , k) and ρ(f ) < σ, then from (2.10) we obtain that the order of growth of h0 is ρ1 = ρ(h0 ) 6 max{ρ, ρ(f )} < σ. By Lemma 2.2, for any ε (0 < ε < σ − ρ1 ) there exists a set E1 ⊂ (1, +∞) with a finite linear measure such that (2.11) |h0 (z)| 6 er ρ1 +ε holds for all z satisfying |z| = r ∈ / [0, 1] ∪ E1 , r → +∞. From the hypotheses of Lemma 2.11, there exists a set H with dens H > 0 (or m(H) = ∞), and there exist positive constants σ > 0, α > 0 such that (2.12) |h0 (z)| > eαr σ holds for all z satisfying |z| = r ∈ H, r → +∞. By (2.11) and (2.12), we conclude σ ρ1 +ε that for all z satisfying |z| = r ∈ H r ([0, 1] ∪ E1 ), r → +∞, we have eαr 6 er and by ε (0 < ε < σ − ρ1 ) this is a contradiction as r → +∞. Consequently, any meromorphic solution f 6≡ 0 of equation (2.9) is transcendental and satisfies ρ(f ) > σ. 3. Proofs of the results Proof of Theorem 1.3. Let f 6≡ 0 be a meromorphic solution of (1.2). It follows from (1.2) that (k) k−1 (j) f X f . (3.1) |A0 | 6 + |Aj | f f j=1 By Lemma 2.11, we know that f is transcendental. By using Lemma 2.1, there is a set E0 ⊂ (0, +∞) having finite linear measure such that for all z satisfying |z| = r ∈ / E0 , we have (j) f (z) 2k (3.2) f (z) 6 r[T (2r, f )] (j = 1, 2, . . . , k). 206 ANDASMAS AND BELAÏDI By Lemma 2.2, for any given ε (0 < ε < σ − ρ) there exists a set E1 ⊂ (1, +∞) with finite linear measure such that |Aj (z)| 6 er (3.3) ρ+ε , j = 1, 2, . . . , k − 1 holds for all z satisfying |z| = r ∈ / [0, 1] ∪ E1 , r → +∞. Also, by the hypotheses of Theorem 1.3, there exists a set H with m(H) = ∞, such that for all z satisfying |z| = r ∈ H, r → +∞, we have σ |A0 (z)| > eαr . (3.4) Hence it follows from (3.1), (3.2), (3.3) and (3.4) that for all z satisfying |z| = r ∈ H r ([0, 1] ∪ E0 ∪ E1 ), r → +∞, we have (3.5) σ eαr 6 r[T (2r, f )]2k + k−1 X er ρ+ε r[T (2r, f )]2k 6 krer ρ+ε [T (2r, f )]2k . j=1 By 0 < ε < σ − ρ, it follows from (3.5) that (3.6) µ(f ) = ρ(f ) = ∞ and ρ2 (f ) > σ. Furthermore, if λ(1/f ) < ∞, then f is a meromorphic solution of (1.2) with ρ(f ) = µ(f ) = ∞, λ(1/f ) < µ(f ) and by Remark 1.2, we have max{ρ(Aj ) : j = 0, 1, . . . , k − 1} = ρ(A0 ) = β < ∞. Thus, by Lemma 2.6, we get (3.7) ρ2 (f ) 6 ρ(A0 ). By (3.6) and (3.7), we conclude that σ 6 ρ2 (f ) 6 ρ(A0 ). Proof of Theorem 1.4. Let f be a meromorphic solution of (1.3). Assume that ρ(f ) < ∞. It follows from (1.3) that (j) (k) k−1 f Ak f X + + . (3.8) |A0 | 6 |A | j f f f j=1 By Lemma 2.11, we know that f is transcendental with ρ(f ) > σ. By the hypothesis λ(1/f ) < σ and Hadamard factorization theorem, we can write f as f (z) = g(z) d(z) , where g(z) and d(z) are entire functions with λ(d) = ρ(d) = λ(1/f ) < σ, ρ(f ) = ρ(g) > σ. By Lemma 2.9, for any given ε (0 < ε < ρ(f )), there exists a set H3 ⊂ [0, +∞) with dens H3 = 1 such that (3.9) M (r, g) > er ρ(g)−ε holds for all z satisfying |z| = r ∈ H3 . By (3.9), for all z satisfying |z| ∈ H3 at which |g(z)| = M (r, g), we get (3.10) |g(z)| > 1. Then, by (3.10), we obtain A (z) d(z)A (z) k k = 6 |d(z)Ak (z)|. f (z) g(z) ON THE GROWTH AND THE ZEROS OF SOLUTIONS 207 We have ρ(d(z)Ak (z)) < σ and set ρ1 = max{ρ(Aj )(j = 1, 2, . . . , k), ρ(d)} < σ, hence by using similar arguments as in the proof of Theorem 1.3, for any given ε (0 < ε < σ − ρ1 ) there exists a set H4 = H ∩ H3 ⊂ [0, +∞) with positive upper density such that for all z satisfying |z| = r ∈ H4 r ([0, 1] ∪ E0 ∪ E1 ), r → +∞, at which |g(z)| = M (r, g), we have (3.13) f (j) (z) 6 r[T (2r, f )]2k (j = 1, 2, . . . , k), f (z) A (z) ρ1 +ε k , 6 er f (z) (3.14) |A0 (z)| > eαr . (3.11) (3.12) |Aj (z)| 6 er ρ1 +ε (j = 1, 2, . . . , k − 1), σ Substituting (3.11)–(3.14) into (3.8), we obtain for all z satisfying |z| = r ∈ H4 r ([0, 1] ∪ E0 ∪ E1 ), r → +∞, at which |g(z)| = M (r, g), that (3.15) e αr σ 2k 6 r[T (2r, f )] + k−1 X r[T (2r, f )]2k + er ρ1 +ε 6 (k + 1)r[T (2r, f )]2k er ρ1 +ε er ρ1 +ε j=1 . Hence by (3.15), we have ρ(f ) = ∞. This is a contradiction which means that the assumption of ρ(f ) < ∞ is not true. Hence, we conclude that ρ(f ) = ∞. Since Ak 6≡ 0, then by Lemma 2.10, we obtain λ(f ) = λ(f ) = ρ(f ) = ∞. By (1.3), it is easy to see that if f has a zero at z0 of order m, m > k, and Aj (j = 0, 1, . . . , k − 1) are analytic at z0 , then Ak must have a zero at z0 of order m − k. Therefore, we get by Ak 6≡ 0 that 1 1 1 k−1 X N r, 6 kN r, + N r, + N (r, Aj ). f f Ak j=0 (3.16) On the other hand, (1.3) may be rewritten as k−1 1 1 f (k) X f (j) Aj = + + A0 . f Ak f f j=1 So, we get (3.17) k 1 k−1 X X 1 f (j) 6 m r, + m(r, Aj ) + m r, + O(1). m r, f Ak f j=0 j=1 208 ANDASMAS AND BELAÏDI By (3.16) and (3.17), we have (3.18) 1 k−1 1 1 X 6 kN r, + N r, + N (r, Aj ) T r, f f Ak j=0 k−1 k X 1 X f (j) + m r, + m(r, Aj ) + m r, + O(1). Ak f j=0 j=1 By (3.18) and the lemma of logarithmic derivative [10], there exists a set E of r of a finite linear measure such that for all r ∈ / E, we have (3.19) k 1 1 X T (r, f ) = T r, + O(1) 6 kN r, + T (r, Aj ) + C log(rT (r, f )), f f j=0 where C is a positive constant. For sufficiently large r, we have (3.20) (3.21) 1 T (r, f ). 2 (j = 0, 1, . . . , k), C log(rT (r, f )) 6 T (r, Aj ) 6 rα+ε where α = max{ρ(Aj ) (j = 0, 1, . . . , k)}. Then, for r ∈ / E sufficiently large, by using (3.20) and (3.21), we conclude from (3.19) that for r ∈ / E and r sufficiently large 1 1 + (k + 1)rα+ε + T (r, f ). T (r, f ) 6 kN r, f 2 So, we get 1 (3.22) T (r, f ) 6 2kN r, + 2(k + 1)rα+ε , r ∈ / E. f Hence, by (3.22), we get ρ2 (f ) 6 λ2 (f ) and then λ2 (f ) > λ2 (f ) > ρ2 (f ). Since by definition, we have λ2 (f ) 6 λ2 (f ) 6 ρ2 (f ), therefore (3.23) λ2 (f ) = λ2 (f ) = ρ2 (f ). Furthermore, if λ(1/f ) < min{µ(f ), σ}, then f is a meromorphic solution of (1.3) with ρ(f ) = ∞, λ(1/f ) < µ(f ) and by Remark 1.2, we have max{ρ(Aj ) : j = 0, 1, . . . , k} = ρ(A0 ) = β < ∞. Thus, by Lemma 2.6, we get (3.24) ρ2 (f ) 6 β. By (3.23) and (3.24), we conclude that λ2 (f ) = λ2 (f ) = ρ2 (f ) 6 ρ(A0 ). Proof of Corollary 1.1. Since A0 is a meromorphic function with finitely many poles and ρ(A0 ) = σ, then by the Hadamard factorization theorem, we can write A0 as A0 (z) = B(z) P (z) , where B(z) is an entire function with ρ(A0 ) = ρ(B) = σ and P (z) is a polynomial. Hence, by Lemma 2.7, for any given ε (0 < ε < σ), there exists a set H1 ⊂ [0, +∞) with dens H1 > 1 − 2σ > 0 (by Proposition 1.1 we have ON THE GROWTH AND THE ZEROS OF SOLUTIONS m(H1 ) = ∞) such that |B(z)| > er Hence σ−ε 209 holds for all z, |z| = r ∈ H1 and r → +∞. σ−ε (3.25) |A0 (z)| = | σ−2ε B(z) er |> > er , m P (z) cr where c > 0 is a constant and m = deg P (z) > 1. Since Aj (j = 0, 1, . . . , k) are meromorphic functions having finitely many poles and since the poles of f can only occur at the poles of Aj (j = 0, 1, . . . , k), then f must have finitely many poles. Therefore, λ(1/f ) = 0. Then, for any given ε with 0 < 2ε < σ − ρ, we have (3.25) and (3.26) ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} < σ − 2ε. Furthermore, all meromorphic solutions f 6≡ 0 of (1.2) or (1.3) satisfy (3.27) λ(1/f ) < σ − 2ε. Then, from (3.25), (3.26) and (3.27) by application of Theorem 1.3 for equation (1.2), we find that every meromorphic solution f 6≡ 0 of (1.2) satisfies µ(f ) = ρ(f ) = ∞ and σ − 2ε 6 ρ2 (f ) 6 σ. By ε (0 < 2ε < σ − ρ) being arbitrary, we obtain ρ2 (f ) = σ. Furthermore, we conclude from Theorem 1.4 that every meromorphic solution f of equation (1.3) satisfies λ(f ) = λ(f ) = ρ(f ) = ∞ and λ2 (f ) = λ2 (f ) = ρ2 (f ) and if λ(1/f ) < µ(f ), then ρ2 (f ) 6 ρ(A0 ). Proof of Corollary 1.2. Since A0 is a meromorphic function with λ( A10 ) < µ(A0 ) 6 ρ(A0 ) = σ < 12 , then by Lemma 2.8, for any ε (0 < ε < σ), there exists a set H2 ⊂ (1, +∞) that has log densH2 > 0 (by Proposition 1.1 we have m(H2 ) = ∞) such that for all z, |z| = r ∈ H2 , we have (3.28) |A0 (z)| > e(1+o(1))r σ−ε . Obviously, the poles of f must occur at the poles of Aj (j = 0, 1, . . . , k), note that all poles of f are of uniformly bounded multiplicity, then by λ(1/A0 ) < σ and ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} < σ, we have λ( f1 ) < σ. Then, for a given ε with 0 < ε < min{σ − ρ, σ − λ( f1 )}, we have (3.28) and (3.29) ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} < σ − ε and λ(1/f ) < σ − ε. Then by (3.28) and (3.29) with application of Theorem 1.3 for equation (1.2), we find that every meromorphic solution f 6≡ 0 whose poles are of uniformly bounded multiplicity of equation (1.2) satisfies µ(f ) = ρ(f ) = ∞ and σ − ε 6 ρ2 (f ) 6 σ. Since ε (0 < ε < min{σ − ρ, σ − λ(1/f )}) being arbitrary, we obtain ρ2 (f ) = σ. Furthermore, with application of Theorem 1.4 for equation (1.3), we find that every meromorphic solution f whose poles are of uniformly bounded multiplicity of equation (1.3) satisfies λ(f ) = λ(f ) = ρ(f ) = ∞, λ2 (f ) = λ2 (f ) = ρ2 (f ) and if λ(1/f ) < µ(f ), then ρ2 (f ) 6 ρ(A0 ). 210 ANDASMAS AND BELAÏDI References 1. B. Belaïdi, Estimation of the hyper-order of entire solutions of complex linear ordinary differential equations whose coefficients are entire functions, Electron. J. Qual. Theory Differ. Equ. 5 (2002), 1–8. , Growth of solutions of certain non-homogeneous linear differential equations with 2. entire coefficients, JIPAM. J. Inequal. Pure Appl. Math. 5(2) (2004), Article 40, 1–9. 3. A. Besicovitch, On integral functions of order < 1, Math. Ann. 97(2) (1927), 677–695. 4. Z. X. 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